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Q. 16

Expert-verified

Geometry

Found in: Page 526

Book edition Student Edition

Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen

Pages 227 pages

ISBN 9780395977279

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Short Answer

Given points A, B, and C. Find AB, BC, and AC. Are A, B, and C collinear?
If so, which point lies between the other two?
16. A(3, 4), B(-3, 0), C(-1, 1)

The values of $A B = 2 \sqrt{13}, B C = \sqrt{5}, A C = 5$ .

A, B, C are not collinear.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step-1 – Given

The given points are $A (3, 4), B (- 3, 0), C (- 1, 1)$ .

Step-2 – To determine

We have to find the lengths AB, BC and AC. Then have to check if A, B, C are collinear. And determine that which point lies between the other two.

Step-3 – Calculation

Using the distance formula, the distance between A and B is:

$\begin{array}{l} A B = \sqrt{{(- 3 - 3)}^{2} + {(0 - 4)}^{2}} \\ A B = \sqrt{{(- 6)}^{2} + {(- 4)}^{2}} \\ A B = \sqrt{36 + 16} \\ A B = \sqrt{52} \\ A B = 2 \sqrt{13} \end{array}$

Using the distance formula, the distance between B and C is:

$\begin{array}{l} B C = \sqrt{{(- 1 - (- 3))}^{2} + {(1 - 0)}^{2}} \\ B C = \sqrt{{(- 1 + 3)}^{2} + {(1 - 0)}^{2}} \\ B C = \sqrt{{(2)}^{2} + {(1)}^{2}} \\ B C = \sqrt{4 + 1} \\ B C = \sqrt{5} \end{array}$

Using the distance formula, the distance between A and C is:

$\begin{array}{l} A C = \sqrt{{(- 1 - 3)}^{2} + {(1 - 4)}^{2}} \\ A C = \sqrt{{(- 4)}^{2} + {(- 3)}^{2}} \\ A C = \sqrt{16 + 9} \\ A C = \sqrt{25} \\ A C = 5 \end{array}$

Note that the sum of any two lines is not equal to length of other line.

So, A, B, C are not collinear.

The values of $A B = 2 \sqrt{13}, B C = \sqrt{5}, A C = 5$ .

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